## College Physics (7th Edition)

(d) $100$
Intensity level of sound is increased from 20 dB to 40 dB. Intensity level $10log\frac{I}{I_{0}}$ $I_{0}=10^{-12}W/m^{2}$ I=sound intensity. For 20 dB, $20=10log\frac{I_{20}}{I_{0}}$ $I_{20}=100I_{0}$ For 40 dB, $40=10log\frac{I_{40}}{I_{0}}$ $I_{40}=10000I_{0}$ Therefore. $\frac{I_{40}}{I_{20}}=100$